The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Explanation

The maximum energy given to X-ray photon is given by
`hv_(max)=eV`
where V=voltage applied to X-ray tube
`v_(max)=(eV)/(h)`
we know that `v_(max)=(c)/(lamda_(min))`or ` l_(min)=(c)/(v_(max))`
`lamda_(min)=(ch)/(eV)`
Here, `lamda-1=(ch)/(eV)` and `lamda_2=(ch)/(eV_2)`
or `(ch)/(e)[(1)/(V_2)-(1)/(V-1)]=lamda_2-lamda_1=trianglelamda=26xx10^-12`
or `(ch)/(e)[(2)/(3V_1)-(1)/(V_1)]=26xx10^-12` `(V_2=1.5V_1)`
Solving for `V_1`, we get `V_1=16000V` or `16kV`